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Grade 12Physical Chemistry

What is the freezing point of an aqua solution that boils at 104 degree Celsius

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8 Years agoGrade 12
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ApprovedApproved Tutor Answer1 Year ago

To determine the freezing point of an aqueous solution that boils at 104 degrees Celsius, we need to consider the properties of solutions and how solutes affect their freezing and boiling points. This phenomenon is known as colligative properties, which depend on the number of solute particles in a solvent rather than the identity of the solute itself.

Understanding Boiling Point Elevation

In a pure water solution, the boiling point is 100 degrees Celsius at standard atmospheric pressure. When a solute is added, it raises the boiling point of the solution. The boiling point elevation can be calculated using the formula:

  • ΔT_b = i * K_b * m

Where:

  • ΔT_b = change in boiling point
  • i = van 't Hoff factor (number of particles the solute breaks into)
  • K_b = ebullioscopic constant of the solvent (for water, it's approximately 0.512 °C kg/mol)
  • m = molality of the solution

In this case, since the boiling point is 104 degrees Celsius, we can find the change in boiling point:

  • ΔT_b = 104 °C - 100 °C = 4 °C

Calculating the Molality

Now, we can rearrange the boiling point elevation formula to solve for molality:

  • m = ΔT_b / (i * K_b)

Assuming the solute does not dissociate (i = 1), we can plug in the values:

  • m = 4 °C / (1 * 0.512 °C kg/mol) ≈ 7.81 mol/kg

Freezing Point Depression

Next, we need to find the freezing point of the solution. The freezing point depression can be calculated using a similar formula:

  • ΔT_f = i * K_f * m

Where:

  • ΔT_f = change in freezing point
  • K_f = cryoscopic constant of the solvent (for water, it's approximately 1.86 °C kg/mol)

Using the previously calculated molality (assuming i = 1 again), we can find the change in freezing point:

  • ΔT_f = 1 * 1.86 °C kg/mol * 7.81 mol/kg ≈ 14.52 °C

Determining the Freezing Point

Since the normal freezing point of pure water is 0 degrees Celsius, we can find the freezing point of the solution:

  • Freezing Point = 0 °C - ΔT_f = 0 °C - 14.52 °C ≈ -14.52 °C

Thus, the freezing point of the aqueous solution that boils at 104 degrees Celsius is approximately -14.52 degrees Celsius. This illustrates how the addition of solutes can significantly alter the physical properties of a solvent, such as its freezing and boiling points.